distribusi binomial, poisson, dan hipergeometrik · distribusi binomial, poisson, dan...

DistribusiBinomial,

Poisson, danHipergeometrik

CHAPTER TOPICS

The Probability of a Discrete Random Variable Covariance and Its Applications in Finance Binomial Distribution Poisson Distribution Hypergeometric Distribution

RANDOM VARIABLE

Random Variable Outcomes of an experiment expressed numerically E.g., Toss a die twice; count the number of times the

number 4 appears (0, 1 or 2 times)

E.g., Toss a coin; assign $10 to head and -$30 to a tail

T = -$30

DISCRETE RANDOMVARIABLE

Discrete Random Variable Obtained by counting (0, 1, 2, 3, etc.) Usually a finite number of different values E.g., Toss a coin 5 times; count the number of

tails (0, 1, 2, 3, 4, or 5 times)

Probability DistributionValues Probability

0 1/4 = .251 2/4 = .502 1/4 = .25

DISCRETE PROBABILITYDISTRIBUTION EXAMPLE

Event: Toss 2 Coins Count # Tails

T T This is using the A Priori ClassicalProbability approach.

DISCRETE PROBABILITYDISTRIBUTION

List of All Possible [Xj , P(Xj) ] Pairs Xj = Value of random variable

P(Xj) = Probability associated with value

Mutually Exclusive (Nothing in Common)

Collective Exhaustive (Nothing Left Out)

0 1 1j jP X P X

SUMMARY MEASURES

Expected Value (The Mean) Weighted average of the probability distribution

E.g., Toss 2 coins, count the number of tails, computeexpected value:

E X X P X

0 .25 1 .5 2 .25 1

SUMMARY MEASURES

Variance Weighted average squared deviation about the

E.g., Toss 2 coins, count number of tails, computevariance:

(continued)

222j jE X X P X

2 2 2 0 1 .2 5 1 1 .5 2 1 .2 5

j jX P X

COVARIANCE AND ITSAPPLICATION

: discrete random variable

: outcome of

: discrete random variable

: outcome of

: probability of occurrence of the

outcome of an

XY i i i ii

X E X Y E Y P X Y

P X Y i

thd the outcome of Yi

COMPUTING THE MEAN FORINVESTMENT RETURNSReturn per $1,000 for two types of investments

100 .2 100 .5 250 .3 $105XE X

200 .2 50 .5 350 .3 $90YE Y

P(Xi) P(Yi) Economic Condition Dow Jones Fund X Growth Stock Y

.2 .2 Recession -$100 -$200

.5 .5 Stable Economy + 100 + 50

.3 .3 Expanding Economy + 250 + 350

Investment

COMPUTING THE VARIANCEFOR INVESTMENT RETURNS

2 2 22 .2 100 105 .5 100 105 .3 250 105

14, 725 121.35X

2 2 22 .2 200 90 .5 50 90 .3 350 90

37,900 194.68Y

P(Xi) P(Yi) Economic Condition Dow Jones Fund X Growth Stock Y

.2 .2 Recession -$100 -$200

.5 .5 Stable Economy + 100 + 50

.3 .3 Expanding Economy + 250 + 350

Investment

COMPUTING THE COVARIANCEFOR INVESTMENT RETURNS

P(XiYi) Economic Condition Dow Jones Fund X Growth Stock Y

.2 Recession -$100 -$200

.5 Stable Economy + 100 + 50

.3 Expanding Economy + 250 + 350

Investment

100 105 200 90 .2 100 105 50 90 .5

250 105 350 90 .3 23,300

The covariance of 23,000 indicates that the two investments arepositively related and will vary together in the same direction.

COMPUTING THE COEFFICIENT OFVARIATION FOR INVESTMENT RETURNS

Investment X appears to have a lower risk (variation)per unit of average payoff (return) than investmentY

Investment X appears to have a higher averagepayoff (return) per unit of variation (risk) thaninvestment Y

121.351.16 116%

194.682.16 216%

SUM OF TWO RANDOMVARIABLES

The expected value of the sum is equal to the sumof the expected values

The variance of the sum is equal to the sum of thevariances plus twice the covariance

The standard deviation is the square root of thevariance

E X Y E X E Y

2 2 2 2X Y X Y XYVar X Y

2X Y X Y

PORTFOLIO EXPECTEDRETURN AND RISK

The portfolio expected return for a two-assetinvestment is equal to the weighted averageof the two assets

Portfolio risk

portion of the portfolio value assigned to asset

E P wE X w E Y

22 2 21 2 1P X Y XYw w w w

COMPUTING THE EXPECTED RETURN ANDRISK OF THE PORTFOLIO INVESTMENT

P(XiYi) Economic Condition Dow Jones Fund X Growth Stock Y

.2 Recession -$100 -$200

.5 Stable Economy + 100 + 50

.3 Expanding Economy + 250 + 350

Investment

Suppose a portfolio consists of an equal investment in each ofX and Y:

0.5 105 0.5 90 97.5E P

2 20.5 14725 0.5 37900 2 0.5 0.5 23300 157.5P

DOING IT IN PHSTAT

PHStat | Decision Making | Covariance and PortfolioAnalysis Fill in the “Number of Outcomes:” Check the “Portfolio Management Analysis” box Fill in the probabilities and outcomes for investment X and Y Manually compute the CV using the formula in the previous

Here is the Excel spreadsheet that contains theresults of the previous investment example:

Microsoft ExcelWorksheet

IMPORTANT DISCRETEPROBABILITY DISTRIBUTIONS

Discrete ProbabilityDistributions

Binomial Hypergeometric Poisson

BINOMIAL PROBABILITYDISTRIBUTION

‘n’ Identical Trials E.g., 15 tosses of a coin; 10 light bulbs taken from

a warehouse 2 Mutually Exclusive Outcomes on Each Trial E.g., Heads or tails in each toss of a coin;

defective or not defective light bulb Trials are Independent The outcome of one trial does not affect the

outcome of the other

BINOMIAL PROBABILITYDISTRIBUTION

Constant Probability for Each Trial E.g., Probability of getting a tail is the same each

time we toss the coin 2 Sampling Methods Infinite population without replacement Finite population with replacement

(continued)

BINOMIAL PROBABILITYDISTRIBUTION FUNCTION

: probability of successes given and

: number of "successes" in sample 0,1, ,

: the probability of each "success"

: sample size

n XXnP X p p

P X X n p

Tails in 2 Tosses of Coin

X P(X)0 1/4 = .25

1 2/4 = .50

2 1/4 = .25

BINOMIAL DISTRIBUTIONCHARACTERISTICS

Mean E.g.,

Variance andStandard Deviation

E X np 5 .1 .5np

n = 5 p = 0.1

0.2.4.6

0 1 2 3 4 5X

1 5 .1 1 .1 .6708np p

BINOMIAL DISTRIBUTION INPHSTAT

PHStat | Probability & Prob. Distributions |Binomial

Example in Excel Spreadsheet

EXAMPLE: BINOMIALDISTRIBUTION

A mid-term exam has 30 multiple choicequestions, each with 5 possible answers. What isthe probability of randomly guessing the answerfor each question and passing the exam (i.e.,having guessed at least 18 questions correctly)?

Are the assumptions for the binomial distribution met?

30 0 .2

18 1 .84245 10

Yes, the assumptions are met.Using results from PHStat:

POISSON DISTRIBUTION

Siméon Poisson

POISSON DISTRIBUTION Discrete events (“successes”) occurring in a given

area of opportunity (“interval”) “Interval” can be time, length, surface area, etc.

The probability of a “success” in a given “interval” isthe same for all the “intervals”

The number of “successes” in one “interval” isindependent of the number of “successes” in other“intervals”

The probability of two or more “successes”occurring in an “interval” approaches zero as the“interval” becomes smaller E.g., # customers arriving in 15 minutes E.g., # defects per case of light bulbs

POISSON PROBABILITYDISTRIBUTION FUNCTION

: probability of "successes" given

: num ber of "successes" per unit

: expected (average) num ber of "successes"

: 2 .71828 (base of natural logs)

XP X X

E.g., Find the probability of 4customers arriving in 3 minuteswhen the mean is 3.6.

3.6 43.6

.19124!

POISSON DISTRIBUTION INPHSTAT

PHStat | Probability & Prob.Distributions | Poisson

POISSON DISTRIBUTIONCHARACTERISTICS

Standard Deviationand Variance

0.2.4.6

0 1 2 3 4 5X

0.2.4.6

0 2 4 6 8 10X

HYPERGEOMETRICDISTRIBUTION

“n” Trials in a Sample Taken from a FinitePopulation of Size N

Sample Taken Without Replacement Trials are Dependent Concerned with Finding the Probability of “X”

Successes in the Sample Where There are “A”Successes in the Population

HYPERGEOMETRICDISTRIBUTION FUNCTION

: probability that successes given , , and

: sam ple size

: population size

: num ber of "successes" in population

: num ber of "successes" in sam ple

0,1, 2,

X n XP X

P X X n N A

E.g., 3 Light bulbs were selectedfrom 10. Of the 10, there were 4defective. What is the probabilitythat 2 of the 3 selected aredefective?

2 12 .30

HYPERGEOMETRIC DISTRIBUTIONCHARACTERISTICS

Variance and Standard Deviation

AE X n

nA N A N n

FinitePopulationCorrectionFactor

HYPERGEOMETRIC DISTRIBUTIONIN PHStat

PHStat | Probability & Prob. Distributions |Hypergeometric …

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